Number of functions from $ \{1, 2, 3, 4, 5, 6\} \to \{1, 2, 3, 4, 5, 6\} $ s. t. $|f^{-1}[\{a\}] |= 1$, $|f^{-1}[\{b\}] |= 2$, $|f^{-1}[\{c\}] |= 3$ Let $ A = \{1, 2, 3, 4, 5, 6\}. \,$ I need to find the number of functions $ \, f : A \to B \, $ such that there are $ \,  a, b, c \in A \, $ for which $ \, \left|f^{-1}[\{a\}]\right| = 1, \left|f^{-1}[\{b\}]\right| = 2 \, $ and $ \, \left|f^{-1}[\{c\}]\right| = 3$.
Did I do it correctly?
First choose $ 3 $ random elements from $ A $, there are $ {6 \choose 3} $ ways to do so. Arrange the elements in an ordered list: $ a, b, c $ - There are $ 3! $ ways to do that.
For each element, choose its $ f^{-1} $.
$ a: {6 \choose 1} = 6$
$ b: \frac{{5 \choose 2}}{2!} = 5$
$ c: $ automatically chosen.
Therefore, according to the rule of product, there are $ {6 \choose 3}\cdot3!\cdot6\cdot5 $ such functions.
 A: Let $a$, $b$ and $c$ be fixed elements of $A$ and consider a function $f$ that satisfies the desired property.
First note that, by the pigeonhole principle, every element in $A$ can only be mapped to either $a$, $b$ or $c$. In other words, $f^{-1}(\{a\}) \cup f^{-1}(\{b\}) \cup f^{-1}(\{c\}) = A$ and the union is disjoint. So it is now a matter of counting partitions of this form.
The preimage of $a$ must have one element, so there are 6 possible choices. The preimage of $b$ must have two elements, and it is also disjoint from the preimage of $a$, hence we must make two choices out of the 5 remaining elements in $A$, i.e. ${5 \choose 2}$. Finally the preimage of $c$ must contain three elements and also be disjoint from the preimages of both $a$ and $b$, hence there is no choice to be made since there are only three elements left.
So, for any choice of $a$, $b$, $c$ there are $6 \cdot {5 \choose 2}$ functions from $A$ to itself that satisfy the required property. There are $3!{6 \choose 3} = \frac{6!}{3!}$ possible picks of $a$, $b$ and $c$, since order does matter here. Hence a total of $3! \cdot {6 \choose 3} \cdot 6 \cdot {5 \choose 2}$ functions that satisfy the desired property.
